# DC Power-Supply

My goal is to eliminate all AA batteries in my project. Can I use one specific DC power supply to power five separate circuits, all currently powered by AA batteries? One of the circuits requires 8V, two of them are at 6V each, one is at 4.5V, and the last one is at 3V. Please let me know if it can be done using existing products, or if I will need to create it using regulators and voltage dividers, etc. Please let me know what voltage DC power-supply I will need for this project.

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How much current do you need at each voltage? – stevenvh Aug 5 '12 at 7:30
Give us some more information about what the circuits are.. I'm guessing the 8V one had 6 AA batteries, 6V one 4 AA, 4.5V 3AA, 3V 2 AA? Do they need to be all powered at the same time? – geometrikal Aug 5 '12 at 7:54
Did you check this question? – AndrejaKo Aug 5 '12 at 7:59

The simplest solution may be to just buy a 9V or 12V AC power adapter.

You didn't mention the currents in your circuits. Without that it's pretty difficult to find an appropriate solution.

You didn't mention if you preferred through-hole components or SMD ones. If you can, I'd say go for SMD for currents < 0.2A because those are much cheaper. If you can't get something like TO-92 for 0.2A or TO-220 for more.

If you work with low currents (low powers) and don't care too much about efficiency you can then obtain lower voltages by using some LDO regulators. You can search the common distributor sites (Mouser, Digikey, Farnell, or more hobbyst sites like Futurlec or maybe Sparkfun) for those:

• 1 x 8V LDO
• 1 x 6V LDO
• 1 x 4.5V LDO

The 4.5V LDO is less common. You can therefore either:

• Use another 6V LDO in series with a diode with a forward voltage of 1.5V or so (or other possibilities)
• Use LM317 and two resistors to get the desired output voltage of 4.5V

Another possible solution may be to use the integrated DC/DC converters. These provide much higher efficiency, though they're (a lot) more expensive. In my opinion for low currents their cost is not justified unless you'd be running on (small) batteries which doesn't seem to be the case.

As for voltage dividers (using resistors) it may not be such a bad idea, expecially for low currents and if your circuits aren't very sensitive to their supply voltage (e.g. they can accept 4.2V instead of 4.5V for instance). It may be the cheapest solution of all though it has a few drawbacks:

• A constant power loss across the two resistors
• Very bad efficiency (since current across the two resistors >> current going to your circuit in order to get a good voltage reference: typically I_resistor > 10 * I_circuit, therefore efficiency < 10% !)
• Limit in current: if for any reasons your circuit demands more current than you designed the resistor bridge for, your current will be limited to a maximum amount determined by the upper resistor of the bridge

I'll post some schematics later if you need them (please let me know).

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You can get DC-DC Buck converters for very few dollars US. (Sorry if that is what user51166 meant by "integrated DC-DC converters.) For example Example Buck Converter. You'll probably want 6, but could power all from a common supply. Since your DC-DC buck converters are going to only be about 90% efficient your total Voltage needed to run those 6 is about 33 Volts (and you'll want a little extra on your common supply to make up for the inefficiency of the Buck converters. A 36 Volt switching power supply is going to cost close to $40 USD or more. You're replacing about 19 AA Batteries. Your cost is about$5.90 USD in bulk so your payback is about $50/$5.90 = 8.4 changes of batteries. Might be worth it if you run those devices often. But the items are going to need to be fairly near the common switch mode supply or your going to have a spider's web of wires running around.

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 This answer is nonsense. A switching power supply by no means has to cost $40, and there's no reason to use a 36-volt supply if the highest required voltage is 8 V. 36 V (or 24 or 48) might make sense as a distribution voltage, but theres nothing that fundamentally requires it. – The Photon Aug 6 '12 at 1:26 By "integrated DC-DC converters" I meant those low-power (small) circuits that TI/ST sell that could configure the output voltage either by an external circuit (for instance TI LM2575) or by microcontroller (for instance TI TPS62350). Those are quite random examples: you can then select the product category at the top of the page on TI site ("Step-Down Regulator") and see all available options. For low power applications they're quite cheap (a few$). You suggested a complete DC-DC converter (for cheap off ebay) though:) I agree that 36V is overkill: I think you meant to cascade them in series – user51166 Aug 6 '12 at 5:34 And I think a series realisation is not even possible: at the input of the (first) voltage regulator goes the + voltage as well as the - (GND) voltage. Assuming you make this happen (how? using a voltage divider?), a series realisation has quite a few drawbacks. In particular reliability (if one fails, all others after it fail), power efficiency (suppose each DC-DC converter has 90% efficiency: the last circuit will have an input voltage efficiency of (90%)^4) and noise (ALL DC-DC converters generate some noise that has to be filtered - but the filtering isn't perfect). – user51166 Aug 6 '12 at 5:40 I apparently caused a misunderstanding about what I was getting at, but however I did indeed introduce an error. What I was suggesting was 6 Buck converters connected in parallel to a common distribution voltage. The error and I think confusion was stating the switched mode power supply have such a high voltage which The Photon correctly pointed out was too high. A 9 to 12 Volt common distribution source would be more than enough assuming adequate current to supply the Buck converters. I never mentioned or intended the Buck converters to be in series and meant they would be in parallel. – rdivilbiss Aug 6 '12 at 23:51